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Jensen's inequality yields the weighted AM-GM inequality

Example

Let $λ_{1}, \dots, λ_{m} \geq 0$ with $\sum_{j} λ_{j} = 1$ and let $a_{1}, \dots, a_{m} > 0$ . Then $\prod_{j = 1}^{m} a_{j}^{λ_{j}} \leq \sum_{j = 1}^{m} λ_{j} a_{j} .$

Facts & Assumptions

Given: Weights $λ_{j} \geq 0$ summing to $1$ and positive numbers $a_{j}$ .

[L1]

Jensen's inequality holds on a probability space (Jensen's integral inequality for a probability measure).

Verification

technique · direct

1.1

Put a discrete probability measure on ${1, \dots, m}$ by[L1, construct] $P ({j}) = λ_{j}$ , let $f (j) = a_{j}$ , and choose the convex function $φ (x) = - \log x$ on $(0, \infty)$ . Applying [L1] gives $- \log ⁣ (\sum_{j = 1}^{m} λ_{j} a_{j}) \leq - \sum_{j = 1}^{m} λ_{j} \log a_{j} .$

2.1

Multiply by $- 1$ and exponentiate to obtain [step 1.1, algebra] ∎ $\prod_{j = 1}^{m} a_{j}^{λ_{j}} \leq \sum_{j = 1}^{m} λ_{j} a_{j},$ the weighted AM-GM inequality.

Depends on

Jensen's integral inequality for a probability measure

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem (7.44) (standard reference, not scraped)